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This is a follow-up to a previous question of mine. I know that any local minimum $x_0$ of a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ has positive semi-definite Hessian $H(x_0) \succeq 0$. If $f$ is analytic, can we say moreover that there exists a neighbourhood $U$ of $x_0$ such that $$ H(x) \succeq 0 $$ for all $x \in U$? I don't see any reason for this being true since positive semi-definiteness is not a continuous property (while positive definiteness is), but can't find a counter-example.

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Here is a very simple counter-example: $$f(x)=x^2y^2.$$ This gives $$H(x,y)=\begin{pmatrix} 2y^2&4xy\\ 4xy&2x^2 \end{pmatrix},$$ in particular $$H(x,x)=2x^2\begin{pmatrix} 1&2\\ 2&1 \end{pmatrix},$$ which is not positive semidefinite for $ x \ne 0$.

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  • $\begingroup$ Simple and elegant answer... Thank you Harald. $\endgroup$ – Nao Jul 12 '18 at 14:53
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    $\begingroup$ You're welcome. And if you should want a function with a strict local minimum, just add a small multiple of $x^4+y^4$. $\endgroup$ – Harald Hanche-Olsen Jul 12 '18 at 16:12

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